1 INTRODUCTION
E X E R C I S E 1.2
S O L U T I O N S
Three definitions of “bit”: (1) A binary digit (p. 1). (2) Past ...
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1 INTRODUCTION
E X E R C I S E 1.2
S O L U T I O N S
Three definitions of “bit”: (1) A binary digit (p. 1). (2) Past tense of “bite” (p. 1). (3) A small amount (pp. 6, 10).
1.3 ASIC Application-Specific Integrated Circuit CAD Computer-Aided Design CD Compact Disc CO Central Office CPLD Complex Programmable Logic Device DAT Digital Audio Tape DIP Dual In-line Pin DVD Digital Versatile Disc FPGA Field-Programmable Gate Array HDL Hardware Description Language IC Integrated Circuit IP Internet Protocol LSI Large-Scale Integration MCM Multichip Module
1–1
2
DIGITAL CIRCUITS
MSI Medium-Scale Integration NRE Nonrecurring Engineering OK Although we use this word hundreds of times a week whether things are OK or not, we have probably rarely wondered about its history. That history is in fact a brief one, the word being first recorded in 1839, though it was no doubt in circulation before then. Much scholarship has been expended on the origins of OK, but Allen Walker Read has conclusively proved that OK is based on a sort of joke. Someone pronounced the phrase “all correct” as “oll (or orl) correct,” and the same person or someone else spelled it “oll korrect,” which abbreviated gives us OK. This term gained wide currency by being used as a political slogan by the 1840 Democratic candidate Martin Van Buren, who was nicknamed Old Kinderhook because he was born in Kinderhook, New York. An editorial of the same year, referring to the receipt of a pin with the slogan O.K., had this comment: “frightful letters . . . significant of the birth-place of Martin Van Buren, old Kinderhook, as also the rallying word of the Democracy of the late election, ‘all correct’ .... Those who wear them should bear in mind that it will require their most strenuous exertions ... to make all things O.K.” [From the American Heritage Electronic Dictionary (AHED), copyright 1992 by Houghton Mifflin Company] PBX Private Branch Exchange PCB
Printed-Circuit Board
PLD Programmable Logic Device PWB Printed-Wiring Board SMT
Surface-Mount Technology
SSI Small-Scale Integration VHDL VHSIC Hardware Description Language VLSI Very Large-Scale Integration 1.4 ABEL Advanced Boolean Equation Language CMOS Complementary Metal-Oxide Semiconductor JPEG Joint Photographic Experts Group MPEG Moving Picture Experts Group OK (see above) PERL According to some, it’s “Practical Extraction and Report Language.” But the relevant Perl FAQ entry, in perlfaq1.pod, says “never write ‘PERL’, because perl isn't really an acronym, apocryphal folklore and post-facto expansions notwithstanding.” (Thanks to Anno Siegel for enlightening me on this.) VHDL VHSIC Hardware Description Language 1.8
In my book, “dice” is the plural of “die.”
2 NUMBER SYSTEMS AND CODES
E X E R C I S E 2.1
(a)
1101011 2 = 6B 16 10110111 2 = B7 16
(b)
(e)
10100.1101 2 = 14.D 16
(f)
(g)
11011001 2 = 331 8 101111.0111 2 = 57.34 8
(h)
(c)
(i) 2.3
(a) (b) (c) (d) (e) (f)
2.5
S O L U T I O N S
(a) (c) (e) (g) (i)
(d)
174003 8 = 1111100000000011 2 67.248 = 110111.0101 2 F3A5 16 = 1111001110100101 2 AB3D 16 = 1010101100111101 2
15C.38 16 = 101011100.00111 2 1023 16 = 1000000100011 2 = 10043 8 7E6A 16 = 111111001101010 2 = 77152 8 (j)
ABCD 16 = 1010101111001101 2 = 125715 8 C350 16 = 1100001101010000 2 = 1415208 9E36.7A 16 = 1001111000110110.0111101 2 = 117066.364 8 DEAD.BEEF 16 = 1101111010101101.1011111011101111 2 = 157255.575674 8 1101011 2 = 107 10 (b) 174003 8 = 63491 10 10110111 2 = 183 10 (d) 67.24 8 = 55.3125 10 10100.1101 2 = 20.8125 10 (f) F3A5 16 = 62373 10 12010 3 = 138 10 7156 8 = 3694 10
(h) (j)
AB3D 16 = 43837 10 15C.38 16 = 348.21875 10
2–1
2–2
2.6
DIGITAL CIRCUITS
(a)
125 10 = 1111101 2 209 10 = 11010001 2
(b)
132 10 = 1000100 2 727 10 = 10402 5
(f)
(g)
(h)
23851 10 = 5D2B 16 57190 10 = DF66 16
(i)
1435 10 = 2633 8
(j)
65113 10 = FE59 16
(c) (e)
2.7
(a)
2.10 (a)
(d)
3489 10 = 6641 8 9714 10 = 22762 8
1100010 110101 + 11001 -------------------------1001110
(b)
1011000 101110 + 100101 -------------------------1010011
(c)
111111110 (d) 11000000 11011101 1110010 + 1100011 + 1101101 --------------------------------------------------------------101000000 11011111
1372 + 4631 -------------------59A3
(b)
4F1A5 + B8D5 ---------------------5AA7A
(c)
F35B + 27E6 --------------------11B41
(d)
1B90F + C44E ---------------------27D5D
2.11 decimal
+ 18
+ 115
+79
–49
–3
–100
signed-magnitude
00010010
01110011
01001111
10110001
10000011
11100100
two’s-magnitude
00010010
01110011
01001111
11001111
11111101
10011100
one’s-complement
00010010
01110011
01001111
11001110
11111100
10011011
2.18
3
hj =
∑ b4j + i ⋅ 2
j
i=0
Therefore, n–1
4n – 1
B =
∑
bi ⋅ 2i =
i–0
∑ hi ⋅ 16i
i=0 n–1
4n – 1
–B =
24 n
–
∑
i=0
bi ⋅
2i
=
16 n
–
∑ h i ⋅ 16 i
i=0
Suppose a 3n-bit number B is represented by an n-digit octal number Q. Then the two’s-complement of B is represented by the 8’s-complement of Q. 2.22 Starting with the arrow pointing at any number, adding a positive number causes overflow if the arrow is advanced through the +7 to – 8 transition. Adding a negative number to any number causes overflow if the arrow is not advanced through the +7 to –8 transition.
EXERCISE SOLUTIONS
2–3
2.24 Let the binary representation of X be x n – 1 x n – 2 …x 1 x 0 . Then we can write the binary representation of Y as x m x m – 1 …x 1 x , where m = n – d . Note that x m – 1 is the sign bit of Y . The value of Y is 0
n–2
Y = –2 m – 1 ⋅ xm – 1 +
∑ xi ⋅ 2 i
i=0
The value of X is n–2
X = – 2 n – 1 ⋅ xn – 1 +
∑ xi ⋅ 2 i
i=0 n–2
∑
= – 2 n – 1 ⋅ xn – 1 + Y + 2 m – 1 ⋅ xm – 1 +
xi ⋅ 2 i
i = m–1 n–2
∑ xi ⋅ 2 i
= – 2 n – 1 ⋅ xn – 1 + Y + 2 ⋅ 2 m – 1 +
i=m
Case 1 ( x m – 1 = 0 ) In this case, X = Y if and only if – 2 n – 1 ⋅ x n – 1 + only if all of the discarded bits ( x m …x n – 1 ) are 0, the same as x m – 1 .
∑i = m x i ⋅ 2 i n–2
= 0 , which is true if and
∑
Case 2 ( x m – 1 = 1 ) In this case, X = Y if and only if – 2 n – 1 ⋅ x n – 1 + 2 ⋅ 2 m – 1 + n – 2 x i ⋅ 2 i = 0 , which i=m is true if and only if all of the discarded bits ( x m …x n – 1 ) are 1, the same as x m – 1 . 2.25 If the radix point is considered to be just to the right of the leftmost bit, then the largest number is 1.11 … 1 and the 2’s complement of D is obtained by subtracting it from 2 (singular possessive). Regardless of the position of the radix point, the 1s’ complement is obtained by subtracting D from the largest number, which has all 1s (plural). 2.28 n–2
B = –b n – 1 ⋅ 2n – 1 +
∑ bi ⋅ 2i
i=0 n–2
2B = – b n – 1 ⋅ 2 n +
∑ bi ⋅ 2 i + 1
i=0
Case 1 ( b n – 1 = 0 ) First term is 0, summation terms have shifted coefficients as specified. Overflow if bn – 2 = 1 . Case 2 ( b n – 1 = 1 ) Split first term into two halves; one half is cancelled by summation term b n – 2 ⋅ 2 n – 1 if b n – 2 = 1 . Remaining half and remaining summation terms have shifted coefficients as specified. Overflow if bn – 2 = 0 . 2.32 001–010, 011–100, 101–110, 111–000. 2.34 Perhaps the designers were worried about what would happen if the aircraft changed altitude in the middle of a transmission. With the Gray code, the codings of “adjacent” alitudes (at 50-foot increments) differ in only one bit. An altitude change during transmission affects only one bit, and whether the changed bit or the original is transmitted, the resulting code represents an altitude within one step (50 feet) of the original. With a binary code, larger altitude errors could result, say if a plane changed from 12,800 feet (0001000000002) to 12,750 feet (0000111111112) in the middle of a transmission, possibly yielding a result of 25,500 feet (0001111111112).
2–4
DIGITAL CIRCUITS
2.37 010
000
011 110
111
100
101 001
3 DIGITAL CIRCUITS
E X E R C I S E 3.1
S O L U T I O N S
The “probably” cases may cause damage to the gate if sustained. (a) 0
(b) 1
(c) 0
(d) undefined
(e) 1
(f) probably 1
(g) probably 0
(h) probably 0
3.3
A logic buffer is a non-linear amplifier that maps the entire set of possible analog input voltages into just two output votages, HIGH and LOW. An audio amplifier has a linear response over its specified operating range, mapping each input voltage into an output voltage that is directly proprtional to the input voltage.
3.6
From the American Heritage Electronic Dictionary (AHED), copyright 1992 by Houghton Mifflin Company: (1) A structure that can be swung, drawn, or lowered to block an entrance or a passageway. (2) a. An opening in a wall or fence for entrance or exit. b. The structure surrounding such an opening, such as the monumental or fortified entrance to a palace or walled city. (3) a. A means of access: the gate to riches. b. A passageway, as in an airport terminal, through which passengers proceed for embarkation. (4) A mountain pass. (5) The total paid attendance or admission receipts at a public event: a good gate at the football game. (6) A device for controlling the passage of water or gas through a dam or conduit. (7) The channel through which molten metal flows into a shaped cavity of a mold. (8) Sports. A passage between two upright poles through which a skier must go in a slalom race. (9) Electronics. A circuit with multiple inputs and one output that is energized only when a designated set of input pulses is received.
3–1
3–2
DIGITAL CIRCUITS
Well, definition (9) is closest to one of the answers that I had in mind. The other answer I was looking for is the gate of a MOS transistor. 3.14 A CMOS inverting gate has fewer transistors than a noninverting one, since an inversion comes “for free.” 3.15 Simple, inverting CMOS gates generally have two transistors per input. Four examples that meet the requirements of the problem are 4-input NAND , 4-input NOR , 2-in, 2-wide AND–OR–INVERT , and 2-in, 2-wide OR– AND–INVERT. 3.18 One way is that a romance could be sparked, and the designers could end up with a lot less time to do their work. Another way is that the stray perfume in the IC production line could have contaminated the circuits used by the designers, leading to marginal operation, more debugging time by the deidicated designers, and less time for romance. By the way, the whole perfume story may be apocryphal. 3.20 Using the maximum output current ratings in both states, the HIGH-state margin is 0.69V and the LOW-state margin is 1.02V. With CMOS loads (output currents less than 20 µA), the margins improve to 1.349V and 1.25V, respectively. 3.21 The first answer for each parameter below assumes commercial operation and that the device is used with the maximum allowable (TTL) load. The number in parentheses, if any, indicates the value obtained under a lesser but specified (CMOS) load. VOHmin
3.84V (4.4V )
VIHmin
3.15V
VILmax
1.35V
VOLmax
0.33V (0.1V)
IImax
1µA
IOLmax
4 mA (20 µA)
IOHmax
-4 mA (-20 µA)
3.22 Current is positive if it flows into a node. Therefore, an output with negative current is sourcing current. 3.23 The 74HC00 output drive is so weak, it’s not good for driving much: Assume that in the LOW state the output pulls down to 0.33V (the maximum V OL spec). Then the output current is ( 5.0V ) ⁄ 120Ω = 41.7mA , which is way more than the 4-mA commercial spec.
(a)
(b) For this problem, you first have to find the Thévenin equivalent of the load, or 148.5Ω in series with 2.25V . In the HIGH state, the gate must pull the output up to 3.84V, a difference of 1.59V across 148.5Ω , requiring 10.7 mA, which is out of spec. In the LOW state, we have a voltage drop of 2.25V – 0.33V across 148.5Ω , so the output must sink 12.9 mA, again out of spec. 3.24 (In the first printing, change “74FCT257T” to “74HC00.”) The specification for the 74HC00 shows a maximum power-supply current of 10 µA when the inputs aree at 0 or 5V, but based on the discussion in Section 3.5.3 we would expect the current to be more when the inputs are at their worst-case values (1.35 or 3.15V). If we consider “nonlogic” input values, the maximum current will flow if the inputs are held right at the switching threshold, approximately VCC/2. 3.26 (In the first printing, change “74FCT257T” to “74HC00.”) Using the formulas on page 119, we can estimate that R p(on) = ( 5.0 – 3.84 ) ⁄ 0.004 = 290Ω or, using the higher value of V OHmin in the spec, that R p(on) = ( 5.0 – 4.4 ) ⁄ 0.00002 = 30KΩ . (The discrepancy shows that the output characteristic of this device is somewhat nonlinear.) We can also estimate R n(on) = 0.33 ⁄ 0.004 = 82.5Ω . 3.29 The purpose of decoupling capacitors is to provide the instantaneous power-supply current that is required during output transitions. Printed-circuit board traces have inductance, which acts as a barrier to current flow at high frequencies (fast transition rates). The farther the capacitor is from the device that needs decoupling, the larger is the instantaneous voltage drop across the connecting signal path, resulting in larger spike (up or down) in the device’s power-supply voltage.
EXERCISE SOLUTIONS
3–3
3.32 (a) 5 ns. 3.38 Smaller resistors result in shorter rise times for LOW-to-HIGH transitions but higher power consumption in the LOW state. Stated another way, larger resistors result in lower power consumption in the LOW state but longer rise times (more ooze) for LOW -to-HIGH transitions. 3.39 The resistor must drop 5.0 – 2.0 – 0.37 = 2.63V with 5mA of current through it. Therefore r = 2.63 ⁄ 0.005 = 526Ω ; a good standard value would be 510Ω . 3.41 (The Secret of the Ooze.) The wired output has only passive pull-up to the HIGH state. Therefore, the time for LOW-to-HIGH transitions, an important component of total delay, depends on the amount of capacitive loading and the size of the pull-up resistor. A moderate capacitive load (say, 100 pF) and even a very strong pull-up resistor (say, 150Ω ) can still lead to time constants and transition times (15 ns in this example) that are longer than the delay of an additional gate with active pull-up. 3.42 The winner is 74FCT-T—48 mA in the LOW state and 15 mA in the HIGH state (see Table 3–8). TTL families don’t come close. 3.46 n diodes are required. 3.49 For each interfacing situation, we compute the fanout in the LOW state by dividing I OLmax of the driving gate by I ILmax of the driven gate. Similarly, the fanout in the HIGH state is computed by dividing IOLmax of the driving gate by I IHmax of the driven gate. The overall fanout is the lower of these two results. Low-state Case 74LS driving 74LS 74LS driving 74S
High-state
Overall
Excess
Ratio
Fanout
Ratio
Fanout
Fanout
State
8mA ---------------0.4mA
20
400µA ----------------20µA
20
20
none
8mA -----------2mA
4
400µA ----------------50µA
8
4
HIGH
Drive
200µA
3.50 For the pull-down, we must have at most a 0.5-V drop in order to create a V IL that is no worse than a standard LS-TTL output driving the input. Since I ILmax = 0.4mA , we get R pd = 0.5 ⁄ 0.004 = 1250Ω and 2 2 ⁄R P pd = V IL pd = ( 0.5 ) ⁄ 1250 = 0.2mW . (Alternatively, P pd = V IL I IL = 0.5 ⋅ 0.0004 = 0.2mW ) For the pull-up, we must have at most a 2.3-V drop in order to create a V IH that is no worse than a standard LS-TTL output driving the input. Since I IHmax = 20µA , we get R pu = 2.3 ⁄ 0.00002 = 115kΩ and 2 ⁄R 2 P pu = V IH pu = ( 2.3 ) ⁄ 115000 = 0.046mW . (Alternatively, we could have calculated the result as P pu = V IHI IH = 2.3 ⋅ 0.00002 = 0.046mW .) The pull-up dissipates less power. 3.52 The main benefit of Schottky diodes is to prevent transistors from saturating, which allows them to switch more quickly. The main drawback is that they raise the collector-to-emitter drop across an almost-saturated transistor, which decreases LOW -state noise margin. 3.55 LOW-state
HIGH-state
R VCC (Ω )
R GND (Ω)
V Thev (V )
R Thev (Ω )
V Thev – V OL (V)
I OL ( mA )
OK?
470 330
— 470
5.0 2.9375
470 193.875
4.5 2.4375
9.57 12.57
no no
V OH – V Thev I OH (V ) ( µA ) 1 MΩ
IOUT = 0 100 pF
1 kΩ + 2.5 V
VOUT
VIN 100 Ω
IOUT
100 pF
1 kΩ + 2.5 V
–
–
At time t = 0 , V OUT is still 4.583 V, but the Thévenin equivalent of the voltage source and the two resistors in the LOW state is 90.9Ω in series with a 0.227-V voltage source. At time t = ∞ , the capacitor will be discharged to the Thévenin-equivalent voltage and V OUT will be 0.227V. In between, the value of V OUT is gov-
EXERCISE SOLUTIONS
3–5
erned by an exponential law: V OUT = 0.227V + ( 4.583 – 0.227V ) ⋅ e = 4.356 ⋅ e – t ⁄ ( 90.9 ⋅ 100 ⋅ 10 = 4.356 ⋅ e ( – t ) ⁄ ( 90.9 ⋅
10 –9 )
–12 )
– t ⁄ ( Rn C L)
V
V
Because of the DC load resistance, the time constant is a little shorter than it was in Section 3.6.1, at 9.09 ns. To obtain the fall time, we must solve the preceding equation for VOUT = 3.5 and VOUT = 1.5, yielding V OUT t = – 9.09 ⋅ 10 – 9 ⋅ ln -----------4.356 t 3.5 = 1.99 ns t 1.5 = 9.69 ns The fall time t f is the difference between these two numbers, or 7.7 ns. This is slightly shorter than the 8.5 ns result in Section 3.6.1 because of the slightly shorter time constant. 3.70 The time constant is 1kΩ ⋅ 50 pF = 50 ns . We solve the rise-time equation for the point at which VOUT is 1.5 V, as on p. 118 of the text: 5.0 – 1.5 t 1.5 = – 50 ⋅ 10 – 9 ⋅ ln --------------------5.0 t 1.5 = 17.83 ns 3.77 The LSB toggles at a rate of 16 MHz. It takes two clock ticks for the LSB to complete one cycle, so the transition frequency is 8 MHz. The MSB’s frequency is 27 times slower, or 62.5 KHz. The LSB’s dynamic power is the most significant, but the sum of the transitions on the higher order bits, a binary series, is equivalent to almost another 8 MHz worth of transitions on a single output bit. Including the LSB, we have almost 16 MHz, but applied to the load capacitance on just a single output. If the different ouputs actually have different load capacitances, then a weighted average would have to be used. 3.81 VCC
(a)
(b) EN A B C D
EN
B
C
L L H H
Q2 OUT
A
D
Q1
(c)
L H L H
H H L L
H L H H
H L L L
Q1 Q2 OUT on off off off
off L on H off Hi-Z off Hi-Z
EN A
OUT
3.84 In the situations shown in the figure, the diode with the lowest cathode voltage is forward biased, and the anode (signal C) is 0.6V higher. However, under the conditions specified in the exercise, neither diode is forward biased, no current flows through R 2 , and V C is 5.0 V.
3–6
DIGITAL CIRCUITS
3.85 /* Transistor parameters */ #define DIODEDROP 0.6 /* volts */ #define BETA 10; #define VCE_SAT 0.2 /* volts */ #define RCE_SAT 50 /* ohms */ #define MAX_LEAK 0.00001 /* amperes */ main() { float Vcc, Vin, R1, R2; float Ib, Ic, Vce;
/* circuit parameters */ /* circuit conditions */
if (Vin < DIODEDROP) { /* cut off */ Ib = 0.0; Ic = Vcc/R2; /* Tentative leakage current, limited by large R2 */ if (Ic > MAX_LEAK) Ic = MAX_LEAK; /* Limited by transistor */ Vce = Vcc - (Ic * R2); } else { /* active or saturated */ Ib = (Vin - DIODEDROP) / R1; if ((Vcc - ((BETA * Ib) * R2)) >= VCE_SAT) { /* active */ Ic = BETA * Ib; Vce = Vcc - (Ic * R2); } else { /* saturated */ Vce = VCE_SAT; Ic = (Vcc - Vce) / (R2 + RCE_SAT); } } }
3.86 In order to turn on Q2 fully, VA must be 1.2V , corresponding to the sum of the base-to-emitter drops of Q2 and Q5. This could happen if both X and Y are 0.95V or higher. In reality, a somewhat higher voltage is required, because the voltage divider consisting of R1, R3, and other components diverts current from the base of Q2 from turning on fully until X and Y are about 1.1V or higher (at 25°C , according to the typical characteristics graphed in the TI TTL Data Book). 3.90 When the output is HIGH, the relay coil will try to pull it to 12 volts. The high voltage will typically cause high current to flow through the output structure back into the 5-V supply and will typically blow up the output. Open-collector TTL outputs theoretically should not have this problem, but most are not designed to withstand the 12-V potential and transistor breakdown will occur. A few TTL open-collector devices are designed with “high-voltage outputs” to solve this problem. 3.92 F = W ⋅ X + Y ⋅ Z W
X
Y
Z
G
F
W
X
Y
Z
G
F
0
0
0
0
1
0
1
0
0
0
1
0
0
0
0
1
1
0
1
0
0
1
1
0
0
0
1
0
1
0
1
0
1
0
1
0
0
0
1
1
0
1
1
0
1
1
0
1
0
1
0
0
1
0
1
1
0
0
0
1
0
1
0
1
1
0
1
1
0
1
0
1
EXERCISE SOLUTIONS
W
X
Y
Z
G
F
W
X
Y
Z
G
F
0
1
1
0
1
0
1
1
1
0
0
1
0
1
1
1
0
1
1
1
1
1
0
1
3–7
3.96 When one module is driving HIGH and the other n – 1 modules are output-disabled, each disabled module has a 74LS125 output sinking 20 µA of leakage current. In addition, each of the n modules has a 74LS04 input sinking 20 µA of leakage current. Thus, the total sink current is ( n – 1 + n ) ⋅ 20 µA . The 74LS125 can source 2.6 mA in the HIGH state, so we find ( n – 1 + n ) ⋅ 20 µA ≤ 2.6mA n ≤ 65 When one module is driving LOW and the other n – 1 modules are output-disabled, each disabled module has a 74LS125 output sourcing 20 µA of leakage current. In addition, each of the n modules has a 74LS04 input sourcing 0.4 mA. Thus, the total source current is ( n – 1 ) ) ⋅ 20 µA + n ⋅ 0.4 µA . The 74LS125 can sink 24 mA in the LOW state, so we find ( n – 1 ) ⋅ 20µA + n ⋅ 0.4mA ≤ 24mA n ≤ 57 Overall, we require n ≤ 57 .
4 COMBINATIONAL LOGIC DESIGN PRINCIPLES
E X E R C I S E
S O L U T I O N S
4.2 T3:
T2: X
X+1
=
1
X
X+X
=
X
0 1
1 1
= =
1 1
0 1
0 1
= =
0 1
X
X⋅X
=
X
0 1
0 1
= =
0 1
X
Y
X+Y
=
Y+X
0 0 1 1
0 1 0 1
0 1 1 1
= = = =
0 1 1 1
4.3 T3′
4.4 T6
4–1
4–2
DIGITAL CIRCUITS
4.5
The original expression assumes precedence of ⋅ over + , that is, the expression is X + ( Y ⋅ Z ) . The parenthesization must be retained for the correct result, X ′ ⋅ ( Y ′ + Z ′ ) , or the precedence must be swapped.
4.6
The answers for parts (a), (b), (c) are as follows. W ⋅ X ⋅ Y ⋅ Z ⋅ (W ⋅ X ⋅ Y ⋅ Z ′ + W ⋅ X ′ ⋅ Y ⋅ Z + W ′ ⋅ X ⋅ Y ⋅ Z + W ⋅ X ⋅ Y ′ ⋅ Z) = W ⋅ X ⋅ Y ⋅ Z ⋅ W ⋅ X ⋅ Y ⋅ Z ′ + W ⋅ X ⋅ Y ⋅ Z ⋅ W ⋅ X ′ ⋅ Y ⋅ Z + W ⋅ X ⋅ Y ⋅ Z ⋅ W ′ ⋅ X ⋅ Y ⋅ Z + W ⋅ X ⋅ Y ⋅ Z ⋅ W ⋅ X ⋅ Y ′ ⋅ Z (T8) = 0 + 0 + 0 + 0 ( T6′, T5′, T2′ ) = 0 ( A4′ )
4.7
(a)
X Y Z
F
0 0 0
(b)
W X Y Z
F
0
0 0 0 0
1
0 0 1
1
0 0 0 1
1
0 1 0
1
0 0 1 0
0
0 1 1
1
0 0 1 1
1
1 0 0
0
0 1 0 0
1
1 0 1
0
0 1 0 1
1
1 1 0
0
0 1 1 0
1
1 1 1
0
0 1 1 1
1
1 0 0 0
1
1 0 0 1
1
1 0 1 0
0
1 0 1 1
1
1 1 0 0
1
1 1 0 1
0
1 1 1 0
0
1 1 1 1
0
4.9 (a) F = X ′ ⋅ Y + Y′ ⋅ X = ( X + Y ) ⋅ ( X ′ + Y ′ ) (b) F = A ⋅ B = ( A + B ) ⋅ ( A + B ′ ) ⋅ ( A ′ + B ) 4.12 (1) Including inverters makes the problem too difficult. (2) In modern PLD-based designs, inverters do cost nothing and really can be ignored. 4.13 X
XY 00
Z
01
11
0 1
10
1 1
1
1
00
YZ W′ ⋅ Y′ ⋅ Z
1
W
WX
X⋅Y
00
01
11
10
1
Z
01
Z
11
1
1
10
1
1
1
W′ ⋅ X X′ ⋅ Y′ ⋅ Z′
1
1 Z
Y
Y
F=X⋅Y+Z (a)
X⋅Y (b)
X F = W′ ⋅ X + X′ ⋅ Y′ ⋅ Z + X ⋅ Y
EXERCISE SOLUTIONS
4–3
4.14 W ⋅ X ⋅ Y′
X′ ⋅ Z′ (a)
X
XY Z
00
01
1
1
0
11
(b) Y′ ⋅ Z′
10 1
01
11
10
1
1
1
00
X′ ⋅ Z′
1
W
WX 00 YZ
01
Z
W ⋅ Y′ ⋅ Z′
1 Z
Y
11
1
1
10
1
1
Y
F′ = X′ ⋅ Z′ + Y′ ⋅ Z′ F = (X + Z) ⋅ (Y + Z)
X′ ⋅ Y
X
F′ = X′ ⋅ Z′ + X ⋅ Y′ + W ⋅ X ⋅ Y′ F = (X + Z) ⋅ (X′ + Y) ⋅ (W′ + X′ + Y)
4.15 (a) Cost is less —one less gate input. X
X + Y′ + Z′
Z′
X XY 00
Z 0
01
0
11
10
0
0
Z
Z
0
1
X′
X′ + Z
X + Y′ + Z′
Y+Z
Y
X′ + Z
Y′
F
Y+Z
Y
4.19 W′ ⋅ Y′ ⋅ Z (a)
WX 00 YZ
W′ ⋅ X′ ⋅ Y′
00
1
W′ ⋅ X′ ⋅ Z
01
1
11
1
W 01
11
10
X′ ⋅ Y′ ⋅ Z′
(b)
WX 00 YZ
d 1
W 01
11
10
00
1
1
01
1
d
11
d
10
1
X′ ⋅ Y′ X′ ⋅ Z Z
Z d
Y
d
1
Y 10
1 W⋅X⋅Y
W⋅Y⋅Z
W′ ⋅ X′
X F = W′ ⋅ Y′ ⋅ Z + W′ ⋅ X′ ⋅ Z + W ⋅ X ⋅ Y
X
F = W′ ⋅ X′ + X′ ⋅ Y′ + X′ ⋅ Z
+ W′ ⋅ X′ ⋅ Y′ or X′ ⋅ Y′ ⋅ Z′
4.20 X′ + Y + Z WX 00 YZ
(a) W + X′ + Z X′ + Y′ + Z′
00
W 01
11
0
10
0
d
0
0
d
0
W′ + Y
(b) YZ
W′ + X
W
WX 00
01
11
00
0
0
01
0
0
0
d
0
0
10
X′ W′ + Y + Z′
W + Y′ + Z′
W + X′ + Y′
01
d
Z 11
0
Z 11
Y
d
Y 10
0
0
0
W + Y′ + Z X
10 X + Y′ + Z W′ + Z
F = (W + X′ + Y′) ⋅ (X′ + Y + Z) ⋅ (W′ + Z′) ⋅ (X + Y′ + Z)
0 W′ + Y′ + Z
X F = X′ ⋅ (W′ + Y′ + Z)
4.21 For the minimal sum-of-products expression to equal the minimal product-of-sums expression, the corresponding maps must have the opposite don’t-cares covered, so that the expressions yield the same value for the don’tcare input combinations. (a) Both mininal sum-of-products expressions cover cell 15; they are equal. The minimal product-of-sums expression also covers cell 15, so the expressions are not equal. The s-of-p and p-of-s expressions require the same number of gates, but the p-of-s requires one fewer input. (b) Both mininal sum-of-products expressions cover cell 3 and 9 and not 15; they are equal. The minimal product-of-sums expression covers cell 15, and not 3 or 9, so the expressions are equal. The p-of-s expression requires fewer gates and inputs.
4–4
DIGITAL CIRCUITS
4.22 Consensus terms that must be added to cover the hazards are “circled” with rectangles. W′ ⋅ Y′
(a)
Y 0
X ⋅ Y′ ⋅ Z
(b) W
WX 00
01
11
1
1
1
1
1 X
W
WX 00
YZ
10
01
11
X ⋅ Y′
00
Y
01
1
1
W⋅X
11
1
1
10
1
1
W ⋅ X′ ⋅ Y′ 10 1
W ⋅ Y′ ⋅ Z
1 Z
Y X⋅Y
X⋅Z
X
4.28 X Y
Q
Analyzing this circuit with the standard method for feedback sequential circuits (Section 5.5), we get the following excitation equation: Q∗ = X ′ ⋅ ( ( Y ⋅ Q )′ + Q ) = X′ ⋅( Y ′ + Q ′ + Q ) = X′ ⋅1 = X ′ Thus, Q∗ is a function of X alone, and is independent of the circuit’s previous “state.”
4.29
X ⋅ 1 = X (T1′) X ⋅ ( Y + Y ′ ) = X (T5) X ⋅ Y + X ⋅ Y ′ = X (T8)
4.31
( X + Y ′ ) ⋅ Y = X ⋅ Y + Y ′ ⋅ Y (T8) = X ⋅ Y + Y ⋅ Y ′ (T6′) = X ⋅ Y + 0 (T5′) = X ⋅ Y (T1)
4.35
X 1 ⋅ X 2 ⋅ … ⋅ X n ⋅ X n = X 1 ⋅ X 2 ⋅ … ⋅ ( X n ⋅ X n ) (T6′, T7′ as required)
= X 1 ⋅ X 2 ⋅ … ⋅ X n (T3′) X 1 + X 2 + … + X n + X n = X 1 + X 2 + … + ( X n + X n ) (T6, T7 as required) = ( X 1 + X 2 + … + Xn ) (T3) 4.37 Figure 3–4(d) is more appropriate, since electrically a TTL NOR gate is just the wired-AND of inverters. 4.39 (a) True. If A ⋅ B = 0 then either A = 0 or B = 0 . If A + B = 1 then either A = 1 or B = 1 . Therefore, A, B = 0, 1 or 1,0 , and A = B ′ .
EXERCISE SOLUTIONS
4.41
4–5
F ( X 1, … ,X i ,X i + 1, …, X n ) = X 1 ′, …, X i ′ ⋅ F ( 0, …, 0, X i + 1, …, X n )
+ X 1 ′, …, X i ′ ⋅ F ( 0, …, 1, X i + 1, …, X n ) …
+ X 1 ′, …, X i ′ ⋅ F ( 1, …, 1, X i + 1, …, X n ) 2 i minterms
2 i combs
A dual theorem may be written based on maxterms. 4.46 Yes, 2-input NAND gates form a complete set of logic gates. We prove the result in the figure on the right by showing that these gates can be used to make 2-input AND gates, 2-input OR gates, and inverters, which form a complete set. 4.52 Take the dual, “multiply out,” and take the dual again. The result is the same as “adding out.” 4.58 (a) 16 ns. (c) 18 ns. (d) 10 ns. 4.61 To make it easier to follow, we’ll take the dual, multiply out, and then take the dual again. Also, we’ll simplify using theorems T3′ and T6′ , otherwise we’ll get a nonminimal result for sure. For Figure 4–27: F = X ⋅Z + Y ′ ⋅Z + X′ ⋅Y ⋅ Z′ FD = ( X + Z) ⋅ ( Y′ + Z )⋅ ( X′ + Y + Z′ )
= X ⋅ Y ′ ⋅ Z ′ + X ⋅ Z ⋅ Y + Z ⋅ Y ′ ⋅ X ′ + Z ⋅ Y ′ ⋅ X ′ + Z ⋅ Z ⋅ X ′ + Z ⋅ Z ⋅ Y (T8, T5′, T2′) = X ⋅ Y ′ ⋅ Z ′ + X ⋅ Y ⋅ Z + X ′ ⋅ Y ′ ⋅ Z ′ + X ′ ⋅ Z + Y ⋅ Z (T3′, T6′) F = ( X + Y ′ + Z ′ ) ⋅ ( X + Y + Z ) ⋅ ( X ′ + Y ′ + Z ) ⋅ ( X ′ + Z ) ⋅ ( Y + Z ) (not minimal)
For Figure 4-29: F = X ⋅Z ′ + Y ′ FD = ( X + Z ′) ⋅ Y ′
= X ⋅ Y ′ + Z ′ ⋅ Y ′ (T8) = X ⋅ Y ′ + Y ′ ⋅ Z ′ (T6′) F = ( X + Y ′ ) ⋅ ( Y ′ + Z ′ ) (minimal)
4.63 W
WX 00
YZ
01
11
10
1
1
W ⋅ Y′ ⋅ Z
00 W′ ⋅ X ⋅ Y
01
Z 1
11
1
Y 10 W′ ⋅ Y ⋅ Z′
1
1
W⋅X⋅Z X
F = W′ ⋅ X ⋅ Y + W′ ⋅ Y ⋅ Z′ + W ⋅ Y′ ⋅ Z + W ⋅ X ⋅ Z
4–6
DIGITAL CIRCUITS
4.69 For part (d), note that it is easiest to work with the product-of-sums directly; rather than multiplying out, one simply enters the 0s on the map. (a)
X
XY 00
Z
01
11
0
(b) CD
1
1
1
1
A
AB
X
10
00
01
11
10
00
1
1
Z
01
1
1
1
1
Z
11 C 10
1
1
1
1
D Y F=X⋅Y+Z
D B F=D
4.70 Note that in these maps are drawn for the “true” function and we’ve written sum terms for the prime implicates (the dual of prime implicants) directly, instead of using the complement method suggested in Section 4.3.6. X+Z (a)
X
XY 00
Z
01
11
0
(b) Y+Z
10
A
AB 00
01
11
10
01
1
1
1
1
11
1
1
1
1
CD
1
00 D
1
1
1
1
1
Z
D Y
C 10 B F=D
4.72 V ⋅ W′ ⋅ Y′ ⋅ Z′
(a) W
WX 00
YZ
01
11
00 01
1
W
WX 10
YZ
V′ ⋅ X ⋅ Z
00
01
1
1
00
1
11
10 W⋅X⋅Z
01
1
1
11
1
1
Z 11
1
Z
1
Y
Y 10
10 V⋅W⋅Z X
X V=1
V=0 F = V′ ⋅ X ⋅ Z + V ⋅ W ⋅ Z + V ⋅ W′ ⋅ Y′ ⋅ Z′
W′ ⋅ X′ ⋅ Y′ ⋅ Z′
V′ ⋅ X′ ⋅ Y′ ⋅ Z′
(b) W
WX 00
YZ 00
01
1
01
W
WX
11
10
1
1
1
1
00
01
11
11
1
1
10
1
1
YZ 00 01 Z
11
1
10
1
Z
1
Y
Y 10
V′ ⋅ W ⋅ Y′ X
X⋅Y⋅Z
V=0 F = V′ ⋅ W ⋅ Y′ + X ⋅ Y ⋅ Z + W′ ⋅ X′ ⋅ Y′ ⋅ Z′ + V ⋅ X ⋅ Y
X V=1
V⋅X⋅Y
EXERCISE SOLUTIONS
4–7
4.73 Note that in these maps are drawn for the “true” function and we’ve written sum terms for the prime implicates (the dual of prime implicants) directly, instead of using the complement method suggested in Section 4.3.6. (a)
V+Z
YZ
W′ + Z
W
WX
W
WX
00
01
11
10
0
0
0
00 01
0
0
11
0
0
10
0
0
W + X + Z′
00
0
W + X + Y′
01
0
0
11
0
0
10
0
00
YZ
01
11
10
0
0
Z
Z
Y
Y 0
0
0
0
0
V+X X X V + W + Y′ V=1 V=0 F = (V + Z) ⋅ (V + X) ⋅ (W′ + Z) ⋅ (V′ + W + Z′) ⋅ (Y′ + Z)
W
WX 00
YZ
01
00
W + Y + Z′ W + X + Z′
V′ + X′ + Y
W + X′ + Y
V + W + X′ + Z
(b)
11
10
01
0
11
0
10
0
00
YZ V′ + X + Z′
0
V′ + W′ + Y
W
WX
0
V′ + W + Z′ Y′ + Z
00
01
11
0
0
0
0
0
0
10
01
0
11
0
0
10
0
0
Z
V′ + Y + Z Z
0
Y
Y 0
0
0 X + Y′
X
V + Y′ + Z
V′ + W′ + X
X
V=0 V=1 F = (X + Y′) ⋅ (V + Y′ + Z) ⋅ (V′ + W′ + Y′) ⋅ (W + X′ + Y′) ⋅ (W + Y + Z′) or (W + X + Z′)
4.74 W
WX
(a)
00
YZ
01
00
01
11
01
1
1
11
1
1
YZ
00 01
W
WX 10
11
10
00 1
1
1
1 Z
Z
11 Y
Y 10
10
U′ ⋅ V′ ⋅ Y′ ⋅ Z
U′ ⋅ V ⋅ X ⋅ Z
X
X
U,V = 0,0 W
WX 00
YZ
01
11
W
WX 10
00
YZ
00 01
F = U′ ⋅ V′ ⋅ Y′ ⋅ Z + U′ ⋅ V ⋅ X ⋅ Z + X ⋅ Y′ ⋅ Z
U,V = 0,1
01
11
1
1
10
00 1
1
01 Z
Z
11
11
Y
Y 10
X ⋅ Y′ ⋅ Z
10 X
X
U,V = 1,0
U,V = 1,1
4.83 The name of the circuit comes from its output equation, F = 2B OR NOT 2B .
2B F tOR
3 t INV
5
74LS138 6 4 5
G1 G2A G2B
1
A 2 B 3
C
E X E R C I S E
Y0 Y1 Y2
15
Y3 Y4 Y5
12
Y6 Y7
9
14 13
11 10
7
COMBINATIONAL LOGIC DESIGN PRINCIPLES
S O L U T I O N S
5.4
READY′ is an expression, with ′ being a unary operator. Use a name like READY_L or /READY instead.
5.8
Both LOW-to-HIGH and HIGH -to-LOW transitions cause positive transitions on the outputs of three gates (every second gate), and negative transitions on the other three. Thus, the total delay in either case is t p = 3t pLH(LS00) + 3t pHL(LS00) = 3 ⋅ 15 + 3 ⋅ 15 = 90 ns Since t pLH and t pHL for a 74LS00 are identical, the same result is obtained using a single worst-case delay of 15 ns.
5.12 The smallest typical delay through one ’LS86 for any set of conditions is 10 ns. Use the rule of thumb, “minimum equals one-fourth to one-third of typical,” we estimate 3 ns as the minimum delay through one gate. Therefore, the minimum delay through the four gates is estimated at 12 ns. The above estimate is conservative, as it does not take into account the actual transitions in the conditions shown. For a LOW -to-HIGH input transition, the four gates have typical delays of 13, 10, 10, and 20 ns, a total of 53 ns, so the minimum is estimated at one-fourth of this or 13 ns. For a HIGH -to-LOW input transition, the four gates have typical delays of 20, 12, 12, and 13 ns, a total of 57 ns, so the minimum is estimated at 14 ns. 5.15 A decoder with active-low outputs ought to be faster, considering that this decoder structure can be implemented directly with inverting gates (which are faster than noninverting) as shown in Figures 5–35 and 5–37. 5.16 The worst-case ’138 output will have a transition in the same direction as the worst-case ’139 output, so we use tpHL numbers for both, which is the worst combination. The delay through the ’139 is 38 ns, and from the
5–1
5–2
DIGITAL CIRCUITS
active-low enable input of the ’138 is 32 ns, for a total delay of 70 ns. Using “worst-case” numbers for the parts and ignoring the structure of the circuit, an overly pessimistic result of 79 ns is obtained. We can also work the problem with 74HCT parts. Worst-case delay through the ’139 is 43 ns, and from the active-low enable input of the ’138 is 42 ns, for a total delay of 85 ns. Ignoring the structure of the circuit, an overly pessimistic result of 88 ns is obtained. We can also work the problem with 74FCT parts. Worst-case delay through the ’139 is 9 ns, and from the active-low enable input of the ’138 is 8 ns, for a total delay of 17 ns. Ignoring the structure of the circuit, a slightly pessimistic result of 18 ns is obtained. Finally, we can work the problem with 74AHCT parts. Worst-case delay through the ’139 is 10.5 ns, and from the active-low enable input of the ’138 is 12 ns, for a total delay of 22.5 ns. Ignoring the structure of the circuit, a slightly pessimistic result of 23.5 ns is obtained. 5.19 (a)
R 6 4 5
Z Y X
(b)
74LS138
+5 V
1
G1 G2A G2B
A B 3 C 2
Y0 Y1 Y2 Y3 Y4 Y5 Y6 Y7
R 6
14 13 12 11
74LS138
+5 V
15
1 2 13
10
4
74LS10 12
5
F
U2
9 7
C B A
1 2 3
G1 G2A G2B A B C
U1
Y0 Y1 Y2 Y3 Y4 Y5 Y6 Y7
15 14 13 12
1 2 13
74LS10 12
F
U2
11 10 9 7
U1
5.21 Both halves of the ’139 are enabled simultaneously when EN_L is asserted. Therefore, two three-state drivers will be enabled to drive SDATA at the same time. Perhaps the designer forgot to put an extra inverter on the signal going to 1G or 2G, which would ensure that exactly one source drives SDATA at all times. 5.22 The total delay is the sum of the decoding delay through the 74LS139, enabling delay of a 74LS151, and delay through a 74LS20: 38 + 30 + 15 = 83 ns . 5.25 The worst-case delay is the sum of the delays through an ’LS280, select-to-output through an ’LS138, and through an ’LS86: 50 + 41 + 30 = 121 ns . 5.30 The worst-case delay is the sum of four numbers: • In U1, the worst-case delay from any input to C4 (22 ns). • In U2, the worst-case delay from C0 to C4 (22 ns). • In U3, the worst-case delay from C0 to C4 (22 ns). • In U4, the worst-case delay from C0 to any sum output (24 ns). Thus, the total worst-case delay is 90 ns. 5.35 With the stated input combination, Y5_L is LOW and the other outputs are HIGH . We have the following cases: (a)
Negating G2A_L or G2B_L causes Y5_L to go HIGH within 18 ns.
(b) Negating G1 causes Y5_L to go HIGH within 26 ns. (c) Changing A or C causes Y5_L to go HIGH within 27 ns (the change propagates through 3 levels of logic internally), and causes Y4_L or Y1_L respectively to go LOW within 41 ns (2 levels). (d) Changing B causes Y5_L to go HIGH within 20 ns (2 levels), and causes Y7_L to go LOW within 39 ns (3 levels). The delays in the ’LS138 are very strange—the worst-case t pHL for 3 levels is shorter than for 2 levels!
EXERCISE SOLUTIONS
5.39 a
D
DC 00
BA 00
01
1
b
11
10
d
1
A + C′
D
DC 00
01
11
10
00
1
1
d
1
01
1
d
1
11
1
d
d
10
1
d
d
BA A′ + B + C′
A′ + B + C + D 01
1
d
1 A
11
1
10
1
1
d
d
d
d
B
A 1
B B
c
C not minimal
D
DC 00
01
11
10
00
1
1
d
1
01
1
1
d
1
11
1
1
d
d
1
d
d
BA
B′ + D′ A + B′ + C′
C not minimal
C′ + D′
A B 10 A + B′ + C
C not minimal
5.46 The inputs are active low and the outputs are active high in this design. I0_L I1_L
Y0
I2_L I3_L Y1
I4_L I5_L I6_L I7_L
I8_L I9_L I10_L I11_L I12_L I13_L I14_L I15_L
Y2
Y3
5–3
5–4
DIGITAL CIRCUITS
5.47 74x04 I7
1
2
U2 I6
3
4
74x04 U2
I5
5
5 5
I4
9
U2
I7_L
4
8
I6_L I5_L I4_L
3
I3_L I2_L I1_L I0_L
13
U2 I3
11
10
U2 I2
13
12
U2 I1
1
2
U3 I0
3
4
U3
6
74x148
6
2 1
12 11 10
A2
U3 EI I7 I6 I5 I4 I3 I2 I1 I0
9 A2 A1
6
A0
9
GS
14
EO
15
8
U3
7
11
10
A0
U3
13
12
U3 U1
A1
IDLE
EXERCISE SOLUTIONS
5.54 An internal logic diagram for the multiplexer is shown below. 1D0
1D1
1D2
2D0
2D1
2D2
3D0
3D1
3D2
4D0
4D1
4D2
5D0
5D1
5D2
S0
S1
(23)
(1)
(2)
1Y
(2)
(3)
(4)
(21)
2Y
(5)
(6)
(7)
(20)
3Y
(8)
(9)
(10)
(19)
4Y
(11)
(18)
(17)
(16)
(13)
(14)
(15)
5Y
5–5
5–6
DIGITAL CIRCUITS
A truth table and pin assignment for the mux are shown below. 74LS998
Inputs
Outputs
13
S1
S0
1Y
2Y
3Y
4Y
5Y
14
0
0
1D0
2D0
3D0
4D0
5D0
23
0
1
1D1
2D1
3D1
4D1
5D1
1
0
1D2
2D2
3D2
4D2
5D2
1
1
0
0
0
0
0
S0
S1 1D0 1 1D1 1D2 2D0
1Y
22
2D1 2D2 6 3D0
2Y
21
3Y
20
4Y
19
5Y
15
2 3 4 5
7 8 9 10 11 18 17 16
3D1 3D2 4D0 4D1 4D2 5D0 5D1 5D2
The mux can be built using a single PLD, a PAL20L8 or GAL20V8; the pin assignment shown above is based on the PLD. The corresponding ABEL program, MUX3BY5.ABL, is shown below. module Mux_3x5 title '5-Bit, 3-Input Multiplexer J. Wakerly, Marquette University' MUX3BY5 device 'P20L8'; " Input pins I1D0, I1D1, I1D2 I2D0, I2D1, I2D2 I3D0, I3D1, I3D2 I4D0, I4D1, I4D2 I5D0, I5D1, I5D2 S0, S1 " Output pins Y1, Y2, Y3, Y4, Y5
pin pin pin pin pin pin
23, 1, 2; 3, 4, 5; 6, 7, 8; 9, 10, 11; 18, 17, 16; 13, 14;
pin 22, 21, 20, 19, 15;
" Set definitions BUS0 = [I1D0,I2D0,I3D0,I4D0,I5D0]; BUS1 = [I1D1,I2D1,I3D1,I4D1,I5D1]; BUS2 = [I1D2,I2D2,I3D2,I4D2,I5D2]; OUT = [Y1, Y2, Y3, Y4, Y5 ]; " Constants SEL0 = ([S1,S0]==[0,0]); SEL1 = ([S1,S0]==[0,1]); SEL2 = ([S1,S0]==[1,0]); IDLE = ([S1,S0]==[1,1]); equations OUT = SEL0 & BUS0 # SEL1 & BUS1 # SEL2 & BUS2 # IDLE & 0; end Mux_3x5
EXERCISE SOLUTIONS
5–7
5.55 This is the actual circuit of a MUX21H 2-input multiplexer cell in LSI Logic’s LCA 10000 series of CMOS gate arrays. When S is 0, the output equals A; when S is 1, the output equals B. 5.60 74x151 7
S0 S1 S2 Ai Bi
11 10 9 4 3 2
Ci
1
EN A B C D0 D1
Y Y
5
Fi 6
D2 D3
15
Di
D4 D5 13 D6 14
12
D7 U1 – U18
5.67 The ’08 has the same pinout as the ’00, but its outputs are the opposite polarity. The change in level at pin 3 of U1 is equivalent to a change at pin 4 of U2 (the input of an XOR tree), which is equivalent in turn to a change at pin 6 of U2 (the parity-generator output). Thus, the circuit simply generated and checked odd parity instead of even. The change in level at pin 6 of U1 changed the active level of the ERROR signal. 5.69 This problem is answered in Section 5.9.3 of the text, which makes it a silly question.
5–8
DIGITAL CIRCUITS
5.75 P Q 74x682
P0 Q0 P1 Q1 P2 Q2 P3 Q3 P4 Q4 P5 Q5 P6 Q6 P7 Q7
2 3 4 5 6 7 8 9 11 12 13 14 15 16 17 18
74x682
P0 Q0 P1 Q1 P2 Q2 P3 Q3 P4 Q4 P5 Q5 P6 Q6 P7 Q7
P EQ Q
P GT Q
19
EQ0_L
GT0_L
1
P16 Q16 P17 Q17 P18 Q18 P19 Q19 P20 Q20 P21 Q21 P22 Q22 P23 Q23
U1
1
74x682
2
P0
3
Q0 P1
4 5 6 7 8 9 11 12 13 14 15 16 17 18
Q1 P2 Q2 P3 Q3 P4 Q4 P5
2 3 4 5 6 7 8 9 11 12 13 14 15 16 17 18
P0 Q0 P1 Q1 P2 Q2 P3 Q3 P4 Q4 P5
19
/EQ2
1
/GT2
U3
74x27 12
PEQQ
13
U4 3
74x27
4
6
5
P EQ Q
19
U4
EQ1_L 74x02 2
1
3
P GT Q
1
U5
GT1_L
11 10 9
74x02
74x27
8 8
U4
9
10
PGTQ
U5
74x02 5
Q5 P6 Q6 P7 Q7
P GT Q
Q5 P6 Q6 P7 Q7
2
P8 Q8 P9 Q9 P10 Q10 P11 Q11 P12 Q12 P13 Q13 P14 Q14 P15 Q15
P EQ Q
6
4
U5
U2
5.79 The function has 65 inputs, and the worst 65-input function (a 65-input parity circuit) has 2 65 – 1 terms in the minimal sum-of-products expression. Our answer can’t be any worse than this, but we can do better. The expression for c 1 has 3 product terms: c 1 = c 0 ⋅ x 0 + c 0 ⋅ y 0 + x 0 ⋅ y 0 The expression for c 2 is c 2 = c 1 ⋅ x 1 + c 1 ⋅ y 1 + x 1 ⋅ y 1 If we substitute our previous expression for c1 in the equation above and “multiply out,” we get a result with 3 + 3 + 1 = 7 product terms. Let us assume that no further reduction is possible. Continuing in this way, we would find that the expression for c 3 has 7 + 7 + 1 = 15 product terms and, in general, the expression for c i has 2 i + 1 – 1 product terms. Thus, the number of terms in a sum-of-products expression for c 32 is no more than 2 33 – 1 , fewer if minimization is possible.
EXERCISE SOLUTIONS
5.80 16-bit group
74S182
MSBs G3 P3
C3 16-bit group G2 P2
C2 16-bit group G1 P1
C1 16-bit group 74LS181
CIN
G P
74S182 G3 P3
G P
C3 74LS181
CIN
G P
G2 P2
C2 74LS181
CIN
G P
G P
G0 P0
G1 P1
C1 74LS181
C0
LSBs
CIN
G P
G0 P0
C0
C0
GOUT POUT
5–9
5–10
DIGITAL CIRCUITS
5.82 74LS00 1
74LS138
+5V R 6 4 5
1
Z Y
G1 G2A G2B A
2
B 3 C
X
3
2
Y0
15
Y1 Y2
14
4
13
5
Y3 Y4 Y5
12
Y6 Y7
9
74LS00
6
74LS00
11
12
10
13
9
F3
U2 8
10
U1
F2
U2 11
74LS00
7
F1
U2
F4
U2
5.91 S0 74x153 14
S1 S2 B[0:3] C[0:3] D[0:3] E[0:3]
2 1 B0
6
C0
5
D0
4
E0
3 15
B1
10
C1
11
D1
12
E1
13
A B 1G 1C0 1C1 1C2 1C3 2G 2C0 2C1 2C2 2C3
1Y
7
BCDE0
74x157 2Y
9
15
BCDE1
1 A0
U1
3 A1
74x153 14 2 1 B2
6
C2
5
D2
4
E2
3 15
B3
10
C3
11
D3
12
E3
A B 1G 1C0 1C1 1C2 1C3 2G 2C0 2C1
2C2 13 2C3
2
5 6
A2
11 10
A3
14 13
1Y
7
2Y
9
G S 1A 1B 2A 2B 3A 3B 4A 4B
T[0:3] 1Y
4
T0
2Y
7
T1
3Y
9
T2
4Y
12 T3
BCDE2 U3
BCDE3
U2
A[0:3]
5.93 The obvious solution is to use a 74FCT682, which has a maximum delay of 11 ns to its PEQQ output. However, there are faster parts in Table 5–3. In particular, the 74FCT151 has a delay of only 9 ns from any select input to Y or Y. To take advantage of this, we use a ’138 to decode the SLOT inputs statically and apply the resulting eight signals to the data inputs of the ’151. By applying GRANT[2–0] to the select inputs of the ’151, we obtain the MATCH_L output (as well as an active-high MATCH, if we need it) in only 9 ns!
7 Sequential Logic Design Principles
E X E R C I S E
S O L U T I O N S
7.2 S R Q
QN
7.3
The latch oscillates if S and R are negated simultaneously. Many simulator programs will exhibit this same behavior when confronted with such input waveforms. S R Q
QN
119
120
DIGITAL CIRCUITS
7.5
J
EN
Q
Q
T
Q
QN
K CLK
7.8
Just tie the J and K inputs together and use as the D input.
7.9
Excitation and output equations:
D1 = Q1 ′ + Q2 D2 = Q2 ′ ⋅ X Z = Q1 + Q2 ′
Excitation/transition table; state/output table: EN
EN
Q1 Q2
0
1
S
0
1
Z
00
10
11
A
C
D
1
10
B
C
C
0
01
C
A
B
1
10
D
C
C
1
01 10 11
10 00 10
Q1* Q2*
7.15 Excitation equations:
S*
D2 = ( Q 1 ⊕ Q0 ) ⊕ ( Q1 ′ ⋅ Q2 ′ ) D1 = Q2 D0 = Q1
Excitation/transition table; state table: Q2 Q1 Q0
Q2* Q1* Q0*
S
S*
000
100
A
E
001
000
B
A
010
101
C
F
011
001
D
B
100
010
101
110
110
111
111
011
7.18 Excitation and output equations: J0 = K0 = EN J1 = K1 = Q0 ⋅ EN MAX = EN ⋅ Q1 ⋅ Q0
E
C
F
G
G
H
H
D
EXERCISE SOLUTIONS
121
Note that the characteristic equation for a J-K flip-flop is Q∗ = J ⋅ Q ′ + K ′ ⋅ Q . Thus, we obtain the following transition equations: Q0∗ = EN ′ ⋅ Q0 + EN ⋅ Q0 ′ Q1∗ = EN ′ ⋅ Q1 + EN ⋅ Q0 ⋅ Q1 ′ + EN ⋅ Q0 ′ ⋅ Q1
Transition/output table; state/output table: EN
EN
Q1 Q2
0
1
S
0
1
00
00,0
01,0
A
A,0
B,0
01
01,0
10,0
B
B,0
C,0
10
10,0
11,0
C
C,0
D,0
11
11,0
00,1
D
D,0
A,1
Q1* Q2*, MAX
S*, MAX
State diagram: EN′
EN′
A
B EN EN
EN
EN
D EN′
C EN′
Timing diagram: CLK EN Q0 Q1 MAX
7.20 This can be done algebraically. If all of the input combinations are covered, the logical sum of the expressions on all the transitions leaving a state must be 1. If the sum is not 1, it is 0 for all input combinations that are uncovered. For double-covered input combinations, we look at all possible pairs of transitions leaving a state. The product of a pair of transition equations is 1 for any double-covered input combinations. (a) State D, Y = 0 is uncovered. (b) State A, (X+Z′) = 0 is uncovered. State B, W = 1 is double-covered; (W+X) = 0 is uncovered. State C, (W+X+Y+Z) = 0 is uncovered; (W ⋅X + W⋅Y + Z⋅Y + Z⋅X) = 1 is double covered. State D, (X⋅Y + ⋅X′⋅Z + W⋅Z) = 0 is uncovered; (W⋅X′⋅Z + W⋅X⋅Y⋅Z) = 1 is double-covered;
122
DIGITAL CIRCUITS
7.21 Table 9–4 on page 804 shows an output-coded state assignment. Here is a corresponding transition list: S
L3Z L2Z L1Z R1Z R2Z R3Z
Transition expression
S*
L3Z* L2Z* L1Z* R1Z* R2Z* R3Z*
IDLE
0
0
0
0
0
0 (LEFT + RIGHT + HAZ)′ IDLE
0
0
0
0
0
0
IDLE
0
0
0
0
0
0 LEFT ⋅ HAZ′ ⋅ RIGHT′
L1
0
0
1
0
0
0
IDLE
0
0
0
0
0
0 HAZ + LEFT ⋅ RIGHT
LR3
1
1
1
1
1
1
IDLE
0
0
0
0
0
0 RIGHT ⋅ HAZ′ ⋅ LEFT′
R1
0
0
0
1
0
0
L1
0
0
1
0
0
0 HAZ′
L2
0
1
1
0
0
0
L1
0
0
1
0
0
0 HAZ
LR3
1
1
1
1
1
1
L2
0
1
1
0
0
0 HAZ′
L3
1
1
1
0
0
0
L2
0
1
1
0
0
0 HAZ
LR3
1
1
1
1
1
1
L3
1
1
1
0
0
0 1
IDLE
0
0
0
0
0
0
R1
0
0
0
1
0
0 HAZ′
R2
0
0
0
1
1
0
R1
0
0
0
1
0
0 HAZ
LR3
1
1
1
1
1
1
R2
0
0
0
1
1
0 HAZ′
R3
0
0
0
1
1
1
R2
0
0
0
1
1
0 HAZ
LR3
1
1
1
1
1
1
R3
0
0
0
1
1
1 1
IDLE
0
0
0
0
0
0
LR3
1
1
1
1
1
1 1
IDLE
0
0
0
0
0
0
The excitation equations and circuit diagram follow directly from this transition list. 7.25 The minimum setup time is the clock period times the duty cycle. That is, the minimum setup time is the time that the clock is 1. 7.27 As shown in Section 7.9.1, the excitation equation for the latch of Figure 7–72 is Y∗ = C ⋅ D + C ′ ⋅ Y + D ⋅ Y Below, we analyze Figure X7.27 in the same way: D C
(C ⋅ D)′
C ⋅ D + (((C ⋅ D)′ ⋅ C) + Y′)′ Y∗
Q
Y
((C ⋅ D)′ ⋅ C)′
The feedback equation is
QN ((C ⋅ D)′ ⋅ C) + Y′
Y∗ = C ⋅ D + ( ( ( C ⋅ D )′ ⋅ C ) + Y ′ )′
= ( C ⋅ D ) + ( ( C ⋅ D )′ ⋅ C )′ ⋅ Y = C ⋅D + ( ( C ⋅ D) + C′ ) ⋅ Y = C ⋅D + ( D + C ′ ) ⋅ Y = C ⋅D + D ⋅Y + C ′ ⋅ Y The feedback equations are the same, and so the circuits have identical steady-state behavior. The circuit in Figure X7.27 is better in two ways. It uses one less gate, and it has one less load on the D input. 7.29 The AND gate in the original circuit is replaced with a NAND gate. As a result, the second flip-flop stores the opposite of the value stored in the original circuit; to compensate, swap connections to its Q and QN outputs.
EXERCISE SOLUTIONS
123
The OR gates in the original circuit are also replaced with NAND gates. As a result, each input must be connected to a signal of the opposite polarity as before, that is, to the complementary flip-flop output. In the case of connections to the second flip-flop, we swapped outputs twice, so the connections remain the same. The final circuit below uses three 2-input NAND gates. X Z
D
Q
D
Q
CLK Q
CLK Q
CLK
7.45 A transition table corresponding to the state table is shown below: AB Q2 Q1 Q0
00
01
11
10
Z
000
001
001
010
010
0
001
011
011
010
010
0
010
001
001
100
100
0
011
011
011
110
010
1
100
001
101
100
100
1
101
011
011
110
010
1
110
001
101
100
100
1
Q2* Q1* Q0*
This table leads to the following Karnaugh maps for the excitation logic, assuming a “minimal cost” treatment of unused states. D0
A
AB 00
01
11
10
00 1
1
0
0
01 1
1
0
0
Q1 Q2
A
AB 00
01
11
10
00 1
1
0
0
01 1
1
0
0
11 d
d
d
d
10 1
1
0
0
Q1 Q2 A′ Q2
11 1
1
0
0
10 1
1
0
0
Q1
Q2 Q1
D2
B
B
Q0=1
01
11
10
00 0
0
0
0
01 0
0
1
1
D1
AB
A
AB
00
01
11
10
00 0
0
1
1
01 0
0
0
0
11 0
0
0
0
10 0
0
0
0
Q1 Q2
00
11
10
00 1
1
1
1
01 1
1
1
1
11 d
d
d
d
10 1
1
1
1
Q2 Q1
B
B
11 0
1
1
1
10 0
1
1
1
11
10
00 0
0
0
0
01 0
0
1
0
11 d
d
d
d
10 0
0
1
0
Q2 Q1
Q0=0
B
Q0=1 Q1 ⋅ A ⋅ B
Q2
Q0=1
01
Q0′ ⋅ Q1 ⋅ A
Q1
Q0=0
Q2 ⋅ A ⋅ B
Q1
01
Q1 Q2
00
Q1 Q2
Q2
Q1′ ⋅ Q2′ ⋅ A A
A
AB
00
Q1 Q2 Q0=0
Q0′ ⋅ Q2 ⋅ A
A
AB
Q0
Q0′ ⋅ Q1 ⋅ B
B
124
DIGITAL CIRCUITS
The resulting excitation equations are D0 = A ′ D1 = Q1 ′ ⋅ Q2 ′ ⋅ A + Q0 D2 = Q2 ⋅ A ⋅ B + Q0 ′ ⋅ Q2 ⋅ A + Q0′ ⋅ Q1 ⋅ A + Q1 ⋅ A ⋅ B + Q0 ′ ⋅ Q1 ⋅ B
Ignoring inverters, a circuit realization with the new equations requires one 2-input gate, six 3-input gates, and one 5-input gate. This is more expensive than Figure 7–54, by four gates. 7.49 The new state assignment yields the following transition/excitation table and Karnaugh maps:
XY Q1 Q0
00
01
11
10
Z
00
00
01
11
01
1
01
01
11
10
11
0
11
11
10
00
10
0
10
10
00
01
00
0
Q2* Q1* or D1 D2 Q1′ ⋅ Q2 ⋅ X Q1′ ⋅ X ⋅ Y
Q1′ ⋅ Q2 ⋅ Y D1
D2
00
01
11
10
00
0
0
1
0
01
0
1
1
1
Q1 Q2
Q1 ⋅ X′ ⋅ Y′
X
XY
Q2′ ⋅ X ⋅ Y′
Q2′ ⋅ X′ ⋅ Y
X
XY 00
01
11
10
00
0
1
0
1
01
1
0
1
0
11
1
0
1
0
0
1
0
1
Q1 Q2
Q2
Q2
11
1
0
0
0
10
1
1
0
1
10
Q1 ⋅ Q2′ ⋅ Y′
Q2 ⋅ X′ ⋅ Y′
Q1
Q1
Y Q1 ⋅ Q2′ ⋅ X′
Y Q2 ⋅ X ⋅ Y
This yields the following excitation equations: D1 = Q1 ′ ⋅ Q2 ⋅ X + Q1 ′ ⋅ Q2 ⋅ Y + Q1 ′ ⋅ X ⋅ Y + Q1 ⋅ Q2 ′ ⋅ X ′ + Q1 ⋅ Q2 ′ ⋅ Y ′ + Q1 ⋅ X ′ ⋅ Y ′ D2 = Q2 ⋅ X ⋅ Y + Q2 ′ ⋅ X ⋅ Y ′ + Q2 ′ ⋅ X ′ ⋅ Y + Q2 ⋅ X ′ ⋅ Y ′
Compared with the results of original state assigment, these equations require two more 3-input AND gates, plus a 6-input OR gate inplace of a 4-input one. However, if we are not restricted to a sum-of-products realization, using the fact that D2 = Q2 ⊕ X ⊕ Y might make this realization less expensive when discrete gates are used.
EXERCISE SOLUTIONS
125
7.57 Here is the transition list: Q2
Q1
Q0
Transition expression
S*
IDLE
0
0
0
(LEFT+RIGHT+HAZ)′
IDLE
0
0
0
IDLE
0
0
IDLE
0
L1
S
Q2*
Q1*
Q0*
IDLE
0
0
0
LEFT
L1
0
0
1
0
HAZ
LR3
1
0
0
0
0
RIGHT
R1
1
0
1
0
0
1
1
L2
0
1
1
L2
0
1
1
1
L3
0
1
0
L3
0
1
0
1
IDLE
0
0
0
R1
1
0
1
1
R2
1
1
1
R2
1
1
1
1
R3
1
1
0
R3
1
1
0
1
IDLE
0
0
0
LR3
1
0
0
1
IDLE
0
0
0
The transition/excitation and output equations below follow directly from the transition list. D2 = Q2∗ = Q2′ ⋅ Q1′ ⋅ Q0′ ⋅ HAZ
+ Q2′ ⋅ Q1′ ⋅ Q0′ ⋅ RIGHT + Q2 ⋅ Q1′ ⋅ Q0 + Q2 ⋅ Q1 ⋅ Q0 = Q2′ ⋅ Q1′ ⋅ Q0′ ⋅ ( HAZ + RIGHT ) + Q2 ⋅ Q0 D1 = Q1∗ = Q2′ ⋅ Q1′ ⋅ Q0
+ Q2′ ⋅ Q1 ⋅ Q0 + Q2 ⋅ Q1′ ⋅ Q0 + Q2 ⋅ Q1⋅ Q0 = Q0 D0 = Q0 ∗ = Q2′ ⋅ Q1′ ⋅ Q0′ ⋅ LEFT
+ Q2′ ⋅ Q1′ ⋅ Q0′ ⋅ RIGHT + Q2′ ⋅ Q1′ ⋅ Q0 + Q2 ⋅ Q1′ ⋅ Q0 = Q2′ ⋅ Q1′ ⋅ Q0′ ⋅ ( LEFT + RIGHT ) + Q1′ ⋅ Q0 Starting from the IDLE state, the following transitions may be observed: Q2
Q1
Q0
LEFT
RIGHT
HAZ
Q2*
Q1*
Q0*
S*
IDLE
0
0
0
1
0
1
1
0
1
R1
IDLE
0
0
0
0
1
1
1
0
1
R1
IDLE
0
0
0
1
1
0
1
0
1
R1
IDLE
0
0
0
1
1
1
1
0
1
R1
S
For each input combination, the machine goes to the R1 state, because R1’s encoding is the logical OR of the encodings of the two or three next states that are specified by the ambiguous state diagram.
126
DIGITAL CIRCUITS
The behavior aboveis not so good and is a result of synthesis choices—state encoding and logic synthesis method. If a different state encoding were used for R1, or if a different synthesis method were used (e.g., product-of-s-terms), then the results could be different. For example, starting with the transition list given earlier, we can obtain the following set of transition equations using the product-of-s-terms method: D2 = Q2∗ = ( Q2 + Q1 + Q0 + LEFT + RIGHT + HAZ )
⋅ ( Q2 + Q1 + Q0 + LEFT′ ) ⋅ ( Q2 + Q1 + Q0′ ) ⋅ ( Q2 + Q1′ + Q0′ ) ⋅ ( Q2 + Q1′ + Q0 ) ⋅ ( Q2′ + Q1′ + Q0 ) ⋅ ( Q2′ + Q1 + Q0 ) = ( Q2 + Q1 + RIGHT + HAZ ) ⋅ ( Q2 + Q1 + LEFT′ ) ⋅ ( Q2 + Q0′ ) ⋅ ( Q1′ + Q0 ) ⋅ ( Q2′ + Q0) D1 = Q1∗ = ( Q2 + Q1 + Q0 + LEFT + RIGHT + HAZ )
⋅ ( Q2 + Q1 + Q0 + LEFT′ ) ⋅ ( Q2 + Q1 + Q0 + HAZ′ ) ⋅ ( Q2 + Q1 + Q0 + RIGHT′ ) ⋅ ( Q2 + Q1′ + Q0 ) ⋅ ( Q2′ + Q1′ + Q0 ) ⋅ ( Q2′ + Q1 + Q0 ) = ( Q2 + Q1 + Q0 ) ⋅ ( Q1′ + Q0 ) ⋅ ( Q2′ + Q0 ) D0 = Q0 ∗ = ( Q2 + Q1 + Q0 + LEFT + RIGHT + HAZ )
⋅ ( Q2 + Q1 + Q0 + HAZ′ ) ⋅ ( Q2 + Q1′ + Q0′ ) ⋅ ( Q2 + Q1′ + Q0 ) ⋅ ( Q2′ + Q1′ + Q0 ′ ) ⋅ ( Q2′ + Q1′ + Q0 ) ⋅ ( Q2′ + Q1 + Q0 ) = ( Q2 + Q0 + LEFT + RIGHT ) ⋅ ( Q2 + Q0 + HAZ′ ) ⋅ ( Q1′ ) ⋅ ( Q2′ + Q0 ) These equations yield the following transitions: Q2
Q1
Q0
LEFT
RIGHT
HAZ
Q2*
Q1*
Q0*
IDLE
0
0
0
0
0
0
0
0
0
IDLE
IDLE
0
0
0
0
1
1
1
0
0
LR3
IDLE
0
0
0
1
1
0
0
0
1
L1
IDLE
0
0
0
1
1
1
0
0
0
IDLE
S
S*
This is obviously different and still not particularly good behavior. 7.58 Let E(SB), E(SC), and E(SD) be the binary encodings of states SB, SC, and SD respectively. Then E ( SD ) = E ( SB ) + E ( SC ) , the bit-by-bit logical OR of E(SB) and E(SC). This is true because the synthesis method uses the logical OR of the next values for each state variable and, by extension, the logical OR of the encoded states.
EXERCISE SOLUTIONS
127
7.68 As far as I know, I was the first person to propose BUT-flops, and Glenn Trewitt was the first person to analyze them, in 1982. To analyze, we break the feedback loops as shown in the figure to the right.
X1 X2
Y1∗
Y1
Y2∗
Y2
Q1 Q2
The excitation and output equations are Y1 = [ ( X1 ⋅ Y1 ) ⋅ ( X2 ⋅ Y2 )′ ]′
= X1 ′ + Y1 ′ + X2 ⋅ Y2 Y2 = [ ( X2 ⋅ Y2 ) ⋅ ( X1 ⋅ Y1 )′ ]′
= X2 ′ + Y2 ′ + X1 ⋅ Y1 Q1 = Y1 Q2 = Y2
The corresponding transition/state table is X1 X2 Y1 Y2
00
01
11
10
00
11
11
11
11
01
11
10
10
11
11
11
10
11
01
10
11
11
01
01
Y1* Y2*
The two stable total states are circled. Notice that state 00 is unreachable. When X1 X2 = 00 or 11, the circuit generally goes to stable state 11, with Q1 Q2 = 11. The apparent oscillation between states 01 and 10 when X1 X2 = 11 may not occur in practice, because it contains a critical race that tends to force the circuit into stable state 11. When X1 X2 = 01 or 10, the Q output corresponding to the HIGH input will oscillate, while the other output remains HIGH . Whether this circuit is useful is a matter of opinion. 7.71 When X=1, the circuit was supposed to “count” through its eight states in Gray-code order. When X=0, it remains in the current state. If this were the case, I suppose it could be used as a 3-bit random number generator. However, I messed up on the logic diagram and the circuit actually does something quite different and completely useless, compared to what I intended when I wrote the problem. Someday I’ll fix this problem. Also, metastability may occur when X is changed from 1 to 0. 7.79 Figure X5.59 requires two “hops” for each input change. Figure 7–66 is faster, requiring only one hop for each input change. On the other hand, Figure 7–66 cannot be generalized for n>2. 7.90 Either this exercise is a joke, or a correct answer is much too dangerous to publish. Nevertheless, Earl Levine offers two possible answers: (Stable output) (Oscillating output)
Was the last answer to this question “yes”? Was the last answer to this question “no”?
8
74x163 2 1 9 7 10 3 4 5 6
CLK CLR LD ENP ENT A B C D
14
QA 13 QB 12
QC 11 QD 15 RCO
E X E R C I S E
Sequential Logic Design Practices
S O L U T I O N S
8.1
In the first three printings, change “RAMBANK0” to “RAMBANK1” in the third line of the exercise. The results are the same. The new expression describes exactly the input combinations in which the 8 high-order bits of ABUS are 000000012, the same as the original expression using don’t-cares.
8.2
The 16-series devices have 64 × 32 or 2048 fuses (see Figure 10–2). The 20-series devices have 64 × 40 or 2560 fuses.
8.3
There are 64 × 32 = 2048 fuses in the AND array (see Figure 10–4). Each of the eight macrocells has one fuse to control the output polarity and one fuse to select registered vs. combinational configuration in the 16V8R, or to assert the output-enable in the 16V8S. There are also two global fuses to select the overall configuration (16V8C, 16V8R, or 16V8S). The total number of fuses is therefore 2048 + 16 + 2 = 2066 . A real 16V8 (depending on the manufacturer) has at least 64 additional fuses to disable individual product terms, 64 user-programmable fuses that do nothing but store a user code, and a security fuse. (Once the security fuse is programmed, the rest of the fuse pattern can no longer be read.)
8.5
The f maxE column below gives the answers in MHz. Part numbers
Suffix
t PD
t CO
t CF
t SU
tH
f maxE
f maxI
PAL16L8, PAL16Rx, PAL20L8, PAL20Rx
-5
5
4
–
4.5
0
117.7
117.7
PAL16L8, PAL16Rx, PAL20L8, PAL20Rx
-7
7.5
6.5
–
7
0
74.1
74.1
PAL16L8, PAL16Rx, PAL20L8, PAL20Rx
-10
10
8
–
10
0
55.6
55.6
GAL22V10
-25
25
15
13
15
0
33.3
35.7
145
146
8.6
DIGITAL CIRCUITS
The f maxI column above gives the answers in MHz.
8.7 module Eight_Bit_Reg title ’8-bit Edge-Triggered Register’ Z74X374 device ’P16V8R’; " Input pins CLK, !OE D1, D2, D3, D4, D5, D6, D7, D8
pin 1, 11; pin 2, 3, 4, 5, 6, 7, 8, 9;
" Output pins Q1, Q2, Q3, Q4, Q5, Q6, Q7, Q8
pin 19, 18, 17, 16, 15, 14, 13, 12;
" Set definitions D = [D1,D2,D3,D4,D5,D6,D7,D8]; Q = [Q1,Q2,Q3,Q4,Q5,Q6,Q7,Q8]; equations Q := D; end Eight_Bit_Reg 8.10 If EN or CLK is 0, the output will be stable. If both are 1, the results are unpredictable, since they depend on circuit timing. It is certain that the circuit’s output will be unstable as long as this condition is true.
8.11 The counter is modified to return to a count of 0 when count 9 is reached. module Z74x162 title '4-bit Decade Counter' "Z74X162 device 'P16V8R'; " Input pins CLK, !OE pin 1, 11; A, B, C, D pin 2, 3, 4, 5; !LD, !CLR, ENP, ENT pin 6, 7, 8, 9; " Output pins QA, QB, QC, QD pin 19, 18, 17, 16 istype 'reg'; RCO pin 15; " Set definitions INPUT = [ D, C, B, A ]; COUNT = [QD, QC, QB, QA ]; equations COUNT.CLK = CLK; COUNT := !CLR & ( LD & INPUT # !LD & (ENT & ENP) & (COUNT < 9) & (COUNT + 1) # !LD & (ENT & ENP) & (COUNT == 9) & 0 # !LD & !(ENT & ENP) & COUNT); RCO = (COUNT == 9) & ENT; end Z74x162 8.13 The counting direction is controlled by QD: count up when QD=1, count down when QD=0. A load occurs when the counter is in the terminal state, 1111 when counting up, 0000 when counting down. The MSB is complemented during a load and the other bits are unchanged. Let us assume that the counter is initially in one of the states 0000–0111. Then the counter counts down (QD=0). Upon reaching state 0000, it loads 1000 and subsequently counts up (QD=1). Upon reaching state 1111, the counter loads 0111, and subsequently counts down, repeating the cycle.
EXERCISE SOLUTIONS
147
If the counter is initially in one of the states 1000–1111, the same cyclic behavior is observed. The counting sequence has a period of 16 and is, in decimal, 8, 9, 10, 11, 12, 13, 14, 15, 7, 6, 5, 4, 3, 2, 1, 0, 8, 9, ... If only the three LSBs are observed, the sequence is 0, 1, 2, 3, 4, 5, 6, 7, 7, 6, 5, 4, 3, 2, 1, 0, 0, 1, ... 8.14 The only difference between a ’163 and a ’161 is that the CLR_L input of ’161 is asynchronous. Thus, the counter will go from state 1010 to state 0000 immediately, before the next clock tick, and go from state 0000 to state 0001 at the clock tick. Observing the state just before each clock tick, it is therefore a modulo-10 counter, with the counting sequence 0, 1, ..., 9, 0, 1, .... Note that this type of operation is not recommended, because the width of the CLR_L pulse is not well controlled. That is, the NAND gate will negate the CLR_L pulse as soon as either one of its inputs goes to 0. If, say, the counter’s QB output clears quickly, and its QD output clears slowly, it is possible for CLR_L to be asserted long enough to clear QB but not QD, resulting in an unexpected next state of 1000, or possibly metastability of the QD output. 8.17 The path from the Q1 counter output (B decoder input) to the Y2_L output has 10 ns more delay than the Q2 and Q0 (C and A) paths. Let us examine the possible Y2_L glitches in Figure 8–43 with this in mind: 3→4 (011→100) Because of the delay in the Q1 path, this transition will actually look like 011→110→100. The Y6_L output will have a 10-ns glitch, but Y2_L will not. 7→0 (111→000) Because of the delay in the Q1 path, this transition will actually look like 111→010→000. The Y2_L output will have a 10-ns glitch, but the others will not. 8.19 The delay calculation is different, depending on the starting state. In the INIT state, U7 and U8 take 21 ns to propagate the CLEAR signal to the outputs. Then U6 requires 20 ns setup time on its D inputs while U3 requires 20 ns setup time on its RIN input. This implies a minimum clock period of 41 ns, assuming zero delay from the control unit. In states M1–M8, the minimum clock period depends on the delay of the combinational logic in the control unit that asserts SELSUM when MPY0 is asserted. However, the most obvious way to do this is to connect MPY0 directly to SELSUM, creating a delay of 0 ns. This assumption is made below. Also, it is assumed that MPY0 is 1 to find the worst case. The figure on the next page shows the worst-case path, in heavy lines, to be 106 ns. Since we would like to use the same clock for all states, the minimum clock period is 106 ns.
148
DIGITAL CIRCUITS
Clock edge
39ns
27ns
U2, U3 outputs valid
U6 Q outputs valid 16ns 16ns
24ns
U4 S outputs valid
U4 C4 output valid
24ns
24ns
14ns
21ns 16ns U5 S outputs valid
U5 C4 output valid
14ns
27ns U7 & U8 outputs valid 27ns
U9 gate output valid 20ns
20ns
20ns U6 D inputs set up
U3 RIN input set up
Ready for next clock
EXERCISE SOLUTIONS
149
8.20 The synchronizer fails if META has not settled by the beginning of the setup-time window for FF2, which is 5 ns before the clock edge. Since the clock period is 40 ns, the available metastability resolution time is 35 ns. The MTBF formula is exp ( t r ⁄ τ ) MTBF ( t r ) = -----------------------T0 ⋅ f ⋅ a Substituting the proper values of τ and T 0 for the ’F74, and of f and a for the problem, we calculate exp ( 35 ⁄ 0.4 ) MTBF ( 35ns ) = ------------------------------------------- ≈ 2 ⋅ 10 28 s 2.0 ⋅ 10 – 4 ⋅ 10 6 ⋅ 10 6 8.22 Refer to the sample data sheet on page 169 of the text: “Not more than one output should be shorted at a time; duration of short-circuit should not exceed one second.” In the switch debounce circuit, the short lasts only for a few tens of nanoseconds, so it’s OK. 8.23 CMOS outputs can “latch up” under certain conditions. According to the Motorola High-Speed CMOS Logic Data book (1988 edition, pp. 4–10), a 74HCT output can latch up if a voltage outside the range – 0.5 ≤ V out ≤ V CC + 0.5V is forced on the output by an external source. In a switch debounce circuit using 74HCT04s, the switch connection to ground is an external source, but the voltage (0 V) is within the acceptable range and should not be a problem. Another potential problem is excessive short-circuit current, but again the data book indicates that shorting the output briefly is not a problem, as long as “the maximum package power dissipation is not violated” (i.e., the short is not maintained for a long time). Similar considerations apply to 74AC and 74ACT devices, but in the balance, such devices are not recommended in the switch-debounce application, as we’ll explain. On one hand, typical 74AC/74ACT devices are even less susceptible to latch-up than 74HCT devices. (For example, see the Motorola FACT Data book, 1988 edition, pp. 2–9.) On the other hand, 74AC/74ACT’s high performance may create noise problems for other devices in a system. In particular, when the 74AC/74ACT HIGH output is shorted to ground, it may momentarily drag the local 5 V power-supply rail down with it, especially if the decoupling capacitors are small, far away, or missing. This will in turn cause incorrect operation of the other, nearby logic devices. 8.25 TTL inputs require significant current, especially in the LOW state. The bus holder cannot supply enough current unless the series resistor is made much smaller, which then creates a significant load on the bus.
150
DIGITAL CIRCUITS
8.26 . library IEEE; use IEEE.std_logic_1164.all; --- Exercise 8-26 -- This code combines the address latch and -- and the decoder and its latch entity latch_decode is port ( abus : in std_logic_vector ( 31 downto 0); avalid : in std_logic; la : out std_logic_vector ( 19 downto 0); romcs, ramcs0, ramcs1, ramcs2 : out std_logic ); end entity latch_decode; architecture behave of latch_decode is
begin process (avalid, abus) begin if (avalid = '1') then la load_shift(1), sdata=>sdata(1)); U3: par2ser port map (clock=>clock, data=>data(23 downto 16), load_shift=>load_shift(2), sdata=>sdata(2)); U4: par2ser port map (clock=>clock, data=>data(31 downto 24), load_shift=>load_shift(3), sdata=>sdata(3)); U5: par2ser port map (clock=>clock, data=>data(39 downto 32), load_shift=>load_shift(4), sdata=>sdata(4)); U6: par2ser port map (clock=>clock, data=>data(47 downto 40), load_shift=>load_shift(5), sdata=>sdata(5)); U7: par2ser port map (clock=>clock, data=>data(55 downto 48), load_shift=>load_shift(6), sdata=>sdata(6)); U8: par2ser port map (clock=>clock, data=>data(63 downto 56), load_shift=>load_shift(7), sdata=>sdata(7)); U9: control port map (clock=>clock, load_shift=>load_shift_master, synch=>synch_master); end wak_8_54_arch;
EXERCISE SOLUTIONS
--************************************* -- Basically an 8-bit shift register -- takes the synch_in signal as an -- input, and outputs an 8 bit signal, -- each consecutive bit delayed by one -- from the previous bit. -- library declaration library IEEE; use IEEE.std_logic_1164.all; use IEEE.std_logic_arith.all; use IEEE.std_logic_unsigned.all; -- top level entity declaration entity shift_synch is port ( clock: in STD_LOGIC; synch_in: in STD_LOGIC; synch: buffer STD_LOGIC_VECTOR (7 downto 0) ); end shift_synch; architecture shift_synch_arch of shift_synch is begin -- low order synch signal is simply passed through -- to output. all others are delayed. synch(0)
